$ g(x) = \int_{-12}^{x}(4t - 7)\,dt\,$ $ g\,'(5)\, =$
Solution: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 4t -7$ is continuous on $[-12,5]$. Applying the theorem We're given: $ g(x) = \int_{-12}^{x}(4t - 7)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =4x - 7$ Evaluating $g'(5)$ $ g'(5)= 4(5) - 7 = 13$ The answer: $g'(5)=13$